the width of a rectangle is six less than the length the area of the rectangle is 58 square feet write and solve and equation to find the dimensions of the rectangle rounded to the nearest tenth
W=L-6
W*L=58
(L-6)*L=58
L^2-6L-58=0
Using the quadratic formula
L=(6+-sqrt(36+232))/2=11.2,5.2 rounded
To find the dimensions of the rectangle, we can set up an equation based on the given information.
Let's represent the length of the rectangle as "L" and the width as "W".
According to the problem, the width of the rectangle is six less than the length, so we have the equation:
W = L - 6
The area of a rectangle is calculated by multiplying its length by its width. So, we can also write the equation for the area of the rectangle as:
Area = L * W
We are given that the area of the rectangle is 58 square feet, so we have:
58 = L * W
Since we know that W = L - 6, we can substitute the value of W in the equation:
58 = L * (L - 6)
Expanding the equation, we have:
58 = L^2 - 6L
Rearranging the equation to standard form, we get:
L^2 - 6L - 58 = 0
Now, we can solve this quadratic equation to find the value of L using factoring, completing the square, or the quadratic formula. After getting the value of L, we can substitute it back into the equation W = L - 6 to find the value of W.
Once we have the values of L and W, we can round them to the nearest tenth as requested.